Problem: Find all values of $k$ so that
\[x^2 - (k - 3) x - k + 6 > 0\]for all $x.$
Solution: If we graph $y = x^2 - (k - 3) x - k + 6,$ then we obtain an upward-facing parabola.  Thus, the inequality
\[x^2 - (k - 3) x - k + 6 > 0\]holds as long as the discriminant of the quadratic is negative.

This gives us
\[(k - 3)^2 - 4(-k + 6) < 0.\]This simplifies to $k^2 - 2k - 15 < 0,$ which factors as $(k + 3)(k - 5) < 0.$  Thus, $k \in \boxed{(-3,5)}.$